Question: $g(n) = -6n^{3}+2n^{2}-4n-1$ $f(x) = 7x^{2}+7x+2(g(x))$ $ g(f(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = 7(1^{2})+(7)(1)+2(g(1))$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = -6(1^{3})+2(1^{2})+(-4)(1)-1$ $g(1) = -9$ That means $f(1) = 7(1^{2})+(7)(1)+(2)(-9)$ $f(1) = -4$ Now we know that $f(1) = -4$ . Let's solve for $g(f(1))$ , which is $g(-4)$ $g(-4) = -6(-4)^{3}+2(-4)^{2}+(-4)(-4)-1$ $g(-4) = 431$